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In basic medi...

Question

In basic medium $C r O_{4}^{2 -}$ oxidizes $S_{2} O_{3}^{2 -}$ to form $S O_{4}^{2 -}$ and itself changes into $C r (O H)_{4}^{-} .$ The volume of $0.154 M C r O_{4}^{2 -}$ required to react with $40 m L$ of $0.25 M S_{2} O_{3}^{2 -}$ is ______ $m L$ .
(Rounded off to the nearest integer)

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Solution

$+ 6 C r O_{4}^{2 -} + {+ 2 S}_{2} O_{3}^{2 -} \to + 3 C r (O H)_{4}^{-} + + 6 S O_{4}^{2 -}$
Accodrding to law of equivalence,
$Milliequivalents of C r O_{4}^{2 -} = Milliequivalents of S_{2} O_{3}^{2 -} 0.154 \times V_{C r O_{4}^{2 -}} \times | 6 - 3 | = 0.25 \times 40 \times | 2 (6 - 2) | ∴ V_{C r O_{4}^{2 -}} = 173 m L$

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Similar questions

Q. In basic medium,

C r O_{4}^{2 -}

S_{2} O_{3}^{2 -}

C r (O H)_{4}^{-}

S O_{4}^{2 -}

. How many mL(nearest integer) of

0.154 M

C r O_{4}^{2 -}

are required to react with

40.0 mL

0.246 M

S_{2} O_{3}^{2 -}

[Hint : 0.0154 M = 0.154 \times 3 N C r O_{4}^{2 -} and 0.246 M = 0.246 \times 8 N S_{2} O_{3}^{2 -}]

Q. In basic solution,

{C r O_{4}}^{2 -}

{S_{2} O_{3}}^{2 -}

{C r (O H)_{4}}^{-}

{S_{2} O_{4}}^{2 -}

. How many mL (nearest integer) of

0.154

{C r O_{4}}^{2 -}

are required to react with

40.0

0.246

{S_{2} O_{3}}^{2 -}

0.0154

= 0.154 \times 3

{C r O_{4}}^{2 -}

0.246

= 0.246 \times 8

{S_{2} O_{3}}^{2 -}

]

Q. In basic medium,

C r O_{4}^{2 -}

S_{2} O_{3}^{2 -}

S O_{4}^{2 -}

and itself changes to

C r (O H)_{4}^{-} .

How many mL of 0.154 M

C r O_{4}^{2 -}

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Q. In basic medium,

C r O_{4}^{2 -}

S_{2} O_{3}^{2 -}

S O_{4}^{2 -}

and itself changes to

C r (O H)_{4}^{-} .

How many mL of 0.154 M

C r O_{4}^{2 -}

are required to react with 40 mL of 0.246 M

S_{2} O_{3}^{2 -}

Q. A solution of

0.1 M K M n O_{4}

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